Show V=Cx iff there are no multiple eigenvalue

[catsarebad]

$L: V\to V$ a diagonalizable linear operator on finite-dim vector space.

show that $V = C_x$ iff there are no multiple eigenvalues

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here $C_x = \operatorname{span} \{x, L(x), L^2(x), \cdots\}$

basically it is a cyclic subspace generated by x that belongs to V.

edit: solved

 

[jvicens]

Hello,

Thank you for your post. I can confirm that your statement is correct. To prove this, we can use the definition of a diagonalizable linear operator. A linear operator $L: V\to V$ is said to be diagonalizable if there exists a basis for $V$ consisting of eigenvectors of $L$. In other words, every vector in the basis is an eigenvector of $L$.

Now, let us assume that $V = C_x$. This means that every vector in $V$ can be written as a linear combination of $x, L(x), L^2(x), \cdots$. Since $x$ is the generator of $C_x$, it must be an eigenvector of $L$. This is because $L(x)$ can be written as a linear combination of $x$ and $L^2(x)$, and so on. Therefore, $x$ is an eigenvector of $L$. Similarly, $L(x)$ is also an eigenvector of $L$, and so on. This means that all the vectors in the basis of $V$ are eigenvectors of $L$, and hence $L$ is diagonalizable.

Conversely, if $L$ is diagonalizable, then every vector in the basis of $V$ is an eigenvector of $L$. This means that $V$ can be expressed as a span of eigenvectors of $L$. Since there are no multiple eigenvalues, each eigenvector is unique. Therefore, the subspace generated by $x$ (which is an eigenvector of $L$) will span all of $V$. Hence, $V = C_x$.

In conclusion, we can say that $V = C_x$ if and only if $L$ is a diagonalizable linear operator with no multiple eigenvalues.

I hope this helps. Let me know if you have any further questions.